The conditions given are:
$$\frac{d}{dt} A(t) = \begin{pmatrix} 2 & 5 \\ 1 & -2 \end{pmatrix}A(t)$$ \begin{equation} {A(0)= I} \end{equation} And using this I am supposed to find $\det(A(t))$ for $t=1,10,100$ I tried taking determinants of both sides, component wise approach but don't seem to be getting anywhere. Would appreciate any push in the right direction.
On
You don’t need to solve the differential equation explicitly in order to solve this problem if you know two key properties of matrices: First, the determinant of a matrix is equal to the product of its eigenvalues, taking multiplicity into account. Second, if $f$ is an analytic function and $\lambda$ an eigenvalue of the matrix $M$, then $f(\lambda)$ is an eigenvalue of $f(M)$. Letting $M$ be the coefficient matrix on the right-hand side of the differential equation, you know from the initial conditions that $A(t)=e^{tM}$, so if $\lambda_1$ and $\lambda_2$ are the eigenvalues of $M$, then the eigenvalues of $A(t)$ are $e^{\lambda_1 t}$ and $e^{\lambda_2 t}$, and so $$\det{A(t)}=e^{\lambda_1 t}e^{\lambda_2 t}=e^{(\lambda_1+\lambda_2)t} = e^{t \operatorname{tr}M}.$$ (Here I used the additional fact that the trace of a matrix is equal to the sum of its eigenvalues.)
In this case, $M$ is traceless, so $\det{A(t)}=1$ for all values of $t$.
If $B=\left(\begin{smallmatrix}2&5\\1&-2\end{smallmatrix}\right)$, then the solution of that differential equation is $A(t)=\exp(tB)$. Now, $(5,1)$ and $(-1,1)$ are eigenvectors of $tB$; the eigenvalues are $3$ and $-3$ respectively. If $M=\left(\begin{smallmatrix}5&-1\\1&1\end{smallmatrix}\right)$ (the columns of $M$ are the eigenvectors of $tB$), then$$M^{-1}.(tB).M=\begin{pmatrix}3t&0\\0&-3t\end{pmatrix}.$$Therefore,$$M^{-1}.e^{tB}.M=\begin{pmatrix}e^{3t}&0\\0&e^{-3t}\end{pmatrix},$$from which it follows that\begin{align}A(t)&=e^{tB}\\&=M.\begin{pmatrix}e^{3t}&0\\0&e^{-3t}\end{pmatrix}.M^{-1}\\&=\frac16\begin{pmatrix}5e^{3t}+e^{-3t}&5e^{3t}-5e^{-3t}\\e^{3t}-e^{-3t}&e^{3t}+5e^{-3t}\end{pmatrix}.\end{align}